A Brief Introduction to Classical and Adelic Algebraic ... - William Stein

156 CHAPTER 20. GLOBAL FIELDS AND ADELES
Proof. The proof proceeds in two steps. First we deduce easily from Lemma 20.1.2
that for almost all v the left hand side of (20.1.2) is contained in the right hand
side. Then we use a trick involving discriminants to show the opposite inclusion for
all but finitely many primes.
Since Ov ⊂ Owi for all i, the left hand side of (20.1.2) is contained in the right
hand side if |ωi| ≤ 1 for 1 ≤ i ≤ n and 1 ≤ j ≤ g. Thus by Lemma 20.1.2, for all
wj
but finitely many v the left hand side of (20.1.2) is contained in the right hand side.
We have just eliminated the finitely many primes corresponding to “denominators”
of some ωi, and now only consider v such that ω1, . . .,ωn ∈ Ow for all w | v.
For any elements a1, . . .,an ∈ Kv ⊗K L, consider the discriminant
D(a1, . . .,an) = Det(Tr(aiaj)) ∈ Kv,
where the trace is induced from the L/K trace. Since each ωi is in each Ow, for
w | v, the traces lie in Ov, so
d = D(ω1, . . .,ωn) ∈ Ov.
Also note that d ∈ K since each ωi is in L. Now suppose that
α =
n
i=1
aiωi ∈ Ow1
⊕ · · · ⊕ Owg,
with ai ∈ Kv. Then by properties of determinants for any m with 1 ≤ m ≤ n, we
have
D(ω1, . . .,ωm−1, α, ωm+1, . . .,ωn) = a 2 mD(ω1, . . .,ωn). (20.1.3)
The left hand side of (20.1.3) is in Ov, so the right hand side is well, i.e.,
a 2 m · d ∈ Ov, (for m = 1, . . .,n),
where d ∈ K. Since ω1, . . .,ωn are a basis for L over K and the trace pairing is
nondegenerate, we have d = 0, so by Theorem 20.1.4 we have |d| v = 1 for all but
finitely many v. Then for all but finitely many v we have that a 2 m ∈ Ov. For these
v, that a 2 m ∈ Ov implies am ∈ Ov since am ∈ Kv, i.e., α is in the left hand side of
(20.1.2).
Example 20.1.7. Let K = Q and L = Q( √ 2). Let ω1 = 1/3 and ω2 = 2 √ 2. In the
first stage of the above proof we would eliminate | · | 3 because ω2 is not integral at
3. The discriminant is
d = D
1
3 , 2√ 2
2 = Det 9 0
=
0 16
32
9 .
As explained in the second part of the proof, as long as v = 2, 3, we have equality
of the left and right hand sides in (20.1.2).